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Question: The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle \(\...

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle θ\theta and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

A

U=KX2U = KX^{2}

B

2gl(1+cos⁡θ)\sqrt{2gl(1 + \cos\theta)}

C

U=KXU = KX

D

a2\frac{a}{2}

Answer

U=KXU = KX

Explanation

Solution

If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position

⇒mgh=12mvmax⁡2⇒v2ghmax⁡\Rightarrow mgh = \frac{1}{2}mv_{\max}^{2} \Rightarrow {v\sqrt{2gh}}_{\max}

Also, from figurecos⁡θ=l−hl\cos\theta = \frac{l - h}{l}

⇒h=l(1−cos⁡θ)\Rightarrow h = l(1 - \cos\theta)

So, v2gl(1−cos⁡θ)max⁡{v\sqrt{2gl(1 - \cos\theta)}}_{\max}